Question

在三角形ABC中a＝1b＝2cosA＝

Answer 1

∵a/cosA=b/(2cosB)=c/(3cosC) 又∵根据正弦定理：a/sinA=b/sinB=c/sinC=2R ∴ sinA/cosA=sinB/(2cosB)=sinC/(3cosC) ∴ tanA=tanB/2=tanC/3＞0，三角形为锐角三角形 ∴tanB=2tanA，tanC=3tanA ∴tanA=tan[π-(B+C)]=-tan(B+C)=(tanB+tanC)/(tanBtanC-1) ∴ tanA=(2tanA+3tanA)/(2tanA*3tanA-1) = 5tanA/(6tan2A-1) ∴ 1=5/(6tan2A-1) ∴tan2A=1 ∴tanA=1 ∴A=45° ∴tanB=2，tanC=3 ∴sinB=tanB/√(1+tan2B)=2√5/5 sinC=tanC/√(1+tan2C)=3√10/10 根据正弦定理：a/sinA=b/sinB=c/sinC b=asinB/sinA，c=asinC/sinA S=1/2bcsinA=3 1/2 * asinB/sinA * asinC/sinA * sinA = 3 a2sinBasinC = 6sinA a2 * 2√5/5 * 3√10/10 = 6*√2/2 a2 * 6/√50 = 6*√2/2 a2 = √100 = 10 a = √10